Find the inverse of

f (x) = - 2 (x+3) ^2 - 1

Step-by-step explanation:

Note that the graph of this function is that of a parabola that opens down and has its vertex at (-3, - 1). If we draw a horiz. line thru this graph, it will intersect the graph in two places below this vertex. Thus, this graph fails the horiz. line test and the function has no inverse.

However, if we restrict x as follows: [-3, ∞)

the graph passes the horiz. line test, and the function has an inverse on this restricted domain.

To find it, do the following:

1) replace "f (x) " with "y": y = - 2 (x+3) ^2 - 1

2) Interchange x and y: x = - 2 (y+3) ² - 1

3) Solve this result for y: 2 (y + 3) ² = - x - 1 ↔ (y + 3) ² = (1/2) (-x - 1)

↔ √ (y + 3) ² = √ (1/2) (-x - 1)

↔ y + 3 = (1/√2) · √ (-[x + 1])

↔ y = - 3 + (1/√2) · √ (-[x + 1])

Note that the domain of the √ function is [0, ∞), so - x - 1 must be ≥ 0.

Simplifying: - x - 1 must be ≥ 0 ↔ - x ≥ 1, or x ≤ 1.

This inverse function is defined ONLY on x ≤ 1 same as (-∞, 1].

4) As a last step, replace this "y" with:

-1

f (x)